GRADUATE QUANTUM MECHANICS: 50 Spring 00 Slutins t Assignment 1. 1. (a) T cnstruct an eigenket f τ a, we take the cmbinatin k e i k r r, (1) where k =(k,k y,k z ). Nw τ k e i k r τ a r e i k r r + a = e i k ( r a) r r = e i k a k. () (b) The actin f H n the state r is H r = E r [ r a + r + a (3) s that the actin f H n k is H k e i k r H r e i k r e i k r E r E [ r a (e i k a + e i k a ) r = E( k) k. (4) where E( k) = E cs( k a) = E (cs k +csk y +csk z ) (5) is the crrespnding energy eigenstate.. (a) Since mmentum peratrs always cmmute, any functin f these peratrs als cmmutes, s that [τ d,τ d =[e i P d/ h,e i P d / h =0 (6) Translatin peratrs cmmute. (b) Rtatins abut different aes d nt cmmute, s that [D(ˆn, φ),d(ˆn,φ ) 0 (7) 1
(c) The inverstin peratr reverses the directin f all translatin, s that πτ d π 1 = τ d (8) Cnsequently, the inversin peratr des nt cmmute with the translatin peratr. [π, τ d 0. (9) (d) Under the inversin peratin, angular mmentum peratrs are invariant, π Jπ 1 = J s that [π, J=0. Cnsequently, the inversin peratin cmmutes with functins f the angular mmentum peratr, and thus cmmutes with the rtatin peratr. [π, D(R) = 0. (10) 3. Sakurai prblem 9. When we time reverse a mmentum eigenstate, we reverse the sign f the mmentum, in additin t cmple cnjugating the state. We therefre epect that the time reversal f φ(p) isφ( p). T shw this eplicitly, ( p Θ α = p Θ ) d D p p φ(p ) = p d D p Θ p φ (p ) = p d D p p φ (p ) = d D p δ (D) (p+p ) {}}{ p p φ (p )=φ ( p) (11) 4. Sakurai prblem 1. We can rewrite the matri as H = AS z + B [S + + S (1) where S ± = S ± is y. Written ut eplicitly fr S =1wehave H A 0 B 0 0 0 (13) B 0 A where I have taken h =1. Takingdet[E1 H =E((E A) B ) we see that the energy eigenvalues are E = A ± B,0 (14) The crrespnding eigenkets are ± = +1 ± 1, (E = A ± B) (15) and fr E =0, 0 = m s =0. The Hamiltnian is invariant under time-reversal, since ΘSΘ 1 = S is unchanged by time-reversal. Since Θ m J =(i) mj m J,wehave Θ ± = ±, Θ 0 = 0, (16) i.e the lwer and upper eigenstates are dd-parity under time reversal, whereas the central state is even-parity under time-reversal.
5. Sakurai, chapter 4, Q 6. This is a tricky prblem. There are tw ways yu culd d it: (i) slving the cmplete prblem but t epnential accuracy r (ii) by directly calculating the matri elements between the states n the left, and right hand side. I shall illustrate methd (ii). T begin, let us cnsider the prblem when the length a is infinitely large. In this case, the wavefunctin fr the left, and right hand grund-states are 0 (>a+ b) ψ R () = ψ R = A sin[k(a + b ) (a <<b) Be κ (<a) 0 (< a b) ψ L () = ψ L = A sin[k(a + b + ) ( b << a) (17) (> a) Be κ m m where κ = (V h + E) V h. Nw the tricky bit is that we need t cnstruct rthgnalized wavefunctins. T d this, we cnstruct ψ R = 1 [ ψ [1 ψ L ψ R 1 R ψ L ψ L ψ R ψ L = ψ L (18) These states are nw rthgnal and nrmalized. We shall nw apprimate the cmplete wavefunctin in the frm ψ = α R ψ R + α L ψ L (19) Applying the Hamiltnian t this epressin, and demanding that H ψ = E ψ, we btain the eigenvalue equatin H ab α b = Eα b,(a, b {R, L}), where [ ψr H H ab ψ R ψ R H ψ L ψ L H ψ R ψ L H ψ. (0) L T evaluate this matri, it is helpful t realize that the cmplete Hamiltnian can be written H = H R + V L = H L + V R (1) where H L is the Hamiltnian fr the left-hand well and H R is the Hamiltnian fr the right-hand well and V R = V [θ( a) θ( a b), V L = V [θ( + a + b) θ( + a), () ψ( ) V () L V Ψ R a V - - -a Fig. 1.: Shwing ψ R () and the ptential V L (). 3
With this set-up, we nte that H L,R ψ L,R = E ψ L,R,whereE is the energy f an islated well. If yu nw cmpute the matri element ψ R H ψ L, yu btain ψ R H ψ L = E ψ R ψ L + ψ R V R ψ L 1 ψl ψ R = ψ R V R ψ L 1 ψl ψ R ψ R V R ψ L. (3) In the last step, we have nted that ψ L ψ R is epnentially smaller than unity, s that terms cntaining this quantity have been drpped. The splitting between the tw states is then ging t be simply ± =± ψ R V R ψ L (4) Nw t calculate this, we need t cmpute the epnential tail in ψ L. Applying cntinuity f the wavefunctin and cntinuity f the lgarithmic derivative, we btain T leading epnential accuracy, this gives A sin kb = Be κa, ktan(kb) = κ (5) A = k = π b b, [ 1+ 1 κb, Carrying ut the integral, we then btain B = b π κb e κa (6) a+b ψ R V R ψ L = V d sin[k(a + b )Be κ a b ( )( ) b = V b κ e κ d sin[ke κ ( )( ) 0 b = V b κ e κ = κb The splitting between the tw levels is then ( V ( b ) 4k V bκ ( h 3 π )( κ e κ(a) mb e κa dime (κ+ik) 0 k /κ ) [ {}} { e ikb Im κ + ik ) e κa (7) where κ = m h V fr large V. E = = h mκb 3 e κa (8) 4
ψ + ψ / R L V - -a a ψ ψ / R L V - -a a Fig..: Shwing the even and dd wavefunctins fr the symmetric ptential well. 5